Linear Equations in Two Variables - Solve a pair of linear equations graphically and interpret intersection meaning

Draw the graph of the linear equation $x + 2y = 6$.

1. Find at least two solutions for the equation.
   • Let $x = 0$: $0 + 2y = 6 \implies 2y = 6 \implies y = 3$. Point is $A(0, 3)$.
   • Let $y = 0$: $x + 2(0) = 6 \implies x = 6$. Point is $B(6, 0)$.
   • Let $x = 2$: $2 + 2y = 6 \implies 2y = 4 \implies y = 2$. Point is $C(2, 2)$.
2. Plot the points $(0, 3)$, $(6, 0)$, and $(2, 2)$ on a Cartesian plane.
3. Use a ruler to draw a straight line passing through these points and extend it with arrows at both ends.

This approach uses the intercept method (setting $x=0$ and $y=0$) to find where the line crosses the axes, which is the most efficient way to graph a linear equation.

Check whether the point $(2, -2)$ lies on the graph of the equation $3x + 4y = -2$.

1. Substitute $x = 2$ and $y = -2$ into the Left Hand Side (LHS) of the equation $3x + 4y = -2$.
2. $LHS = 3(2) + 4(-2) = 6 - 8 = -2$.
3. $RHS = -2$.
4. Since $LHS = RHS$, the point $(2, -2)$ satisfies the equation.

If a point satisfies the algebraic equation of the line, it must geometrically lie on the line when represented on a graph. Since the calculation holds true, the point $(2, -2)$ is a part of the line's graph.

1. For $x + y = 3$, points are $(0, 3)$ and $(3, 0)$.
2. For $x - y = 1$, points are $(1, 0)$ and $(0, -1)$.
3. Plotting these lines, they intersect at $(2, 1)$.
4. Verification: $2 + 1 = 3$ and $2 - 1 = 1$. The solution is $x = 2, y = 1$.

To solve graphically, we find at least two points for each line, draw the lines on the coordinate plane, and find the coordinates where they cross.

1. The line $y = x$ passes through $(0, 0)$ and $(4, 4)$.
2. The line $x = 4$ is a vertical line passing through $x = 4$.
3. The $x$-axis is the line $y = 0$.
4. The vertices of the triangle are $(0, 0)$, $(4, 0)$, and $(4, 4)$.
5. Base $b = 4$ units (from $x=0$ to $x=4$ on the $x$-axis).
6. Height $h = 4$ units (from $y=0$ to $y=4$ at $x=4$).
7. Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$ sq units.

Intersection points of the three lines define the vertices of the triangle. The area is then calculated using the geometric formula for a triangle.